Question

the grade point averages (gpa) for 12 randomly selected college students are shown on the right. complete parts (a) through (c) below. assume the population is normally distributed. 2.1 3.1 2.7 1.5 0.9 4.0 2.1 1.1 3.5 0.3 2.4 3.3 (a) find the sample mean. \\( \overline { x } = \square \\) (round to two decimal places as needed.) (b) find the sample standard deviation. \\( s = \square \\) (round to two decimal places as needed.) (c) construct a 95\\% confidence interval for the population mean \\( \mu \\). a 95\\% confidence interval for the population mean is (round to two decimal places as needed.)

Explanation:

Step1: Calculate the sample mean

The formula for the sample mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$.
Here, $n = 12$, and $x_{i}$ are the data points: $2.1,3.1,2.7,1.5,0.9,4.0,2.1,1.1,3.5,0.3,2.4,3.3$.
$\sum_{i=1}^{12}x_{i}=2.1 + 3.1+2.7+1.5+0.9+4.0+2.1+1.1+3.5+0.3+2.4+3.3=29$.
$\bar{x}=\frac{29}{12}\approx2.42$.

Step2: Calculate the sample standard deviation

The formula for the sample standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}$.
First, calculate $(x_{i}-\bar{x})^{2}$ for each $x_{i}$:
For $x = 2.1$: $(2.1 - 2.42)^{2}=(- 0.32)^{2}=0.1024$.
For $x = 3.1$: $(3.1 - 2.42)^{2}=(0.68)^{2}=0.4624$.
For $x = 2.7$: $(2.7 - 2.42)^{2}=(0.28)^{2}=0.0784$.
For $x = 1.5$: $(1.5 - 2.42)^{2}=(-0.92)^{2}=0.8464$.
For $x = 0.9$: $(0.9 - 2.42)^{2}=(-1.52)^{2}=2.3104$.
For $x = 4.0$: $(4.0 - 2.42)^{2}=(1.58)^{2}=2.4964$.
For $x = 2.1$: $(2.1 - 2.42)^{2}=(-0.32)^{2}=0.1024$.
For $x = 1.1$: $(1.1 - 2.42)^{2}=(-1.32)^{2}=1.7424$.
For $x = 3.5$: $(3.5 - 2.42)^{2}=(1.08)^{2}=1.1664$.
For $x = 0.3$: $(0.3 - 2.42)^{2}=(-2.12)^{2}=4.4944$.
For $x = 2.4$: $(2.4 - 2.42)^{2}=(-0.02)^{2}=0.0004$.
For $x = 3.3$: $(3.3 - 2.42)^{2}=(0.88)^{2}=0.7744$.
$\sum_{i = 1}^{12}(x_{i}-\bar{x})^{2}=0.1024 + 0.4624+0.0784+0.8464+2.3104+2.4964+0.1024+1.7424+1.1664+4.4944+0.0004+0.7744 =14.576$.
$s=\sqrt{\frac{14.576}{12 - 1}}=\sqrt{\frac{14.576}{11}}\approx1.15$.

Step3: Construct the 95% confidence interval

Since the population standard deviation $\sigma$ is unknown and $n<30$, we use the $t$-distribution. The degrees of freedom $df=n - 1=11$. For a 95% confidence interval, the critical value $t_{\alpha/2}$ (from the $t$-table) is $t_{0.025,11}=2.201$.
The formula for the confidence interval is $\bar{x}\pm t_{\alpha/2}\frac{s}{\sqrt{n}}$.
$\frac{s}{\sqrt{n}}=\frac{1.15}{\sqrt{12}}\approx0.33$.
The lower limit: $\bar{x}-t_{\alpha/2}\frac{s}{\sqrt{n}}=2.42-2.201\times0.33=2.42 - 0.73=1.69$.
The upper limit: $\bar{x}+t_{\alpha/2}\frac{s}{\sqrt{n}}=2.42+2.201\times0.33=2.42 + 0.73=3.15$.

Answer:

(a) $\bar{x}\approx2.42$
(b) $s\approx1.15$
(c) A 95% confidence interval for the population mean is $(1.69,3.15)$